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This is an exercise for the book Abstract Algebra by Dummit and Foote (pg. 530): Find the degree of $\alpha:=1+\sqrt[3]{2}+\sqrt[3]{4}$ over $\mathbb{Q}$

My efforts:

I first try to find the minimal polynomial by writing $\alpha=1+\sqrt[3]{2}+\sqrt[3]{4}\implies\alpha-1=\sqrt[3]{2}(1+\sqrt[3]{2})\implies(\alpha-1)^{3}=2(1+\sqrt[3]{2})^{3}$ but I didn't manage to get the minimal polynomial from this (which is, according to Wolfram, of degree $3$).

I also tried another method that failed: I noted that $\mathbb{Q}(\alpha)\subset\mathbb{Q}(\sqrt[3]{2})$ hence is of degree $\leq3$, moreover, since it is a subfield and $3$ is prime it only remains to show that $\alpha$ is not rational (which I can't prove).

Can someone pleases help me show that the degree is $3$ (preferably is one of the two methods I tried) ?

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Would this help? $$\eqalign{ & {\left( {x - 1} \right)^3} = 2{\left( {\root 3 \of 2 + 1} \right)^3} = 2\left( {2 + 3\root 3 \of 2 + 3\root 3 \of 4 + 1} \right) = 2\left( {3 + 3\left( {x - 1} \right)} \right) = 6x \cr & {\left( {x - 1} \right)^3} - 6x = 0 \cr} $$ –  Pedro Tamaroff Sep 9 '12 at 17:47
    
@PeterTamaroff - Is $x=\alpha$ ? –  Belgi Sep 9 '12 at 17:56
    
Yes, of course. –  Pedro Tamaroff Sep 9 '12 at 17:59
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5 Answers

up vote 11 down vote accepted

If $1+ \sqrt[3]{2} + \sqrt[3]{4}=r$ with $r$ rational, then $\sqrt[3]{2}$ would satisfy $x^2+x+1-r=0$, contradicting the fact that is has degree $3.$

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How did you get this equality ? (I can't verify it) –  Belgi Sep 9 '12 at 17:49
    
@Belgi Have you tried putting $x=\sqrt[3]{2}$ into $x^2+x+1-r$ and seeing what you get? –  Ragib Zaman Sep 9 '12 at 17:52
    
Thanks, I understand what you did –  Belgi Sep 9 '12 at 17:54
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Let $\beta=\sqrt[3]2$. Then $\alpha=1+\beta+\beta^2$, hence $\mathbb Q(\alpha)\subseteq \mathbb Q(\beta)$. Clearly, the minimal polynomial of $\beta$ is $X^3-2$, hence $[\mathbb Q(\beta):\mathbb Q]=3$. As a vector space, $\mathbb Q(\beta)$ has $1, \beta, \beta^2$ as a basis, hence $\alpha$ is irrational (the only way to write $\alpha=a+b\beta+c\beta^2$ with rational coefficients is $\alpha=1+\beta+\beta^2$). Thus $1<[\mathbb Q(\alpha):\mathbb Q]|3$, i.e. and $[\mathbb Q(\alpha):\mathbb Q]=3$.


You can find the minimal polynomial of $\alpha$ itself: $$\alpha = 1+\beta+\beta^2$$ $$\alpha^2 = (1+\beta+\beta^2)^2=1+2\beta+3\beta^2+2\beta^3+\beta^4 = 5+4\beta+3\beta^2$$ $$\alpha^3 = (1+\beta+\beta^2)(5+4\beta+3\beta^2)=5+9\beta+12\beta^2+7\beta^3+3\beta^4=19+15\beta+12\beta^2$$ Find a combination that eliminates all $\beta$ and $\beta^2$: $$\alpha^2-3\alpha=2+\beta$$ $$\alpha^3-4\alpha^2=-1-\beta$$ $$\Rightarrow\quad \alpha^3-3\alpha^2-3\alpha-1=0$$

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How did you deduce that $\alpha$ is irrational ? –  Belgi Sep 9 '12 at 17:46
    
So its degree is 2*3=6 right ? –  mick Sep 9 '12 at 17:51
    
@Belgi: I clarified this ($1,\beta,\beta^2$ are linearly independant). –  Hagen von Eitzen Sep 9 '12 at 17:59
    
You can type faster than I can Hagen! –  Mark Bennet Sep 9 '12 at 17:59
    
@mick: No the degree of $\mathbb Q(\alpha)=\mathbb Q(\beta)$ is 3 because we have only added the real roots, not the two complex roots. –  Hagen von Eitzen Sep 9 '12 at 17:59
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Your second method is the right one. To see that $\alpha$ is not rational, note that $$\frac{a}{b}=\alpha\implies \frac{a-b}{b\sqrt[3]{2}}=(1+\sqrt[3]{2})\implies \frac{(a-b)^3}{2b^3}=3\alpha\implies(a-b)^3=6ab^2$$ and note that this last equation is homogeneous, so if it has a solution it in $\mathbb Z$ it has a solution with $a$ and $b$ coprime. But $6|(a-b)^3$ so $6|(a-b)$, and so $6^2|ab^2$ so either $6|a$ or $6|b$, but either way since $6|(a-b)$ we get $6|a$ and $6|b$, hence $a$ and $b$ are not coprime. Thus no solution exists, so $\alpha$ is irrational.

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Denote $\alpha = 1 + \sqrt[3]{2} + \sqrt[3]{4}$. Since $\alpha \in \mathbb{Q}(\sqrt[3]{2})$, we get $\mathbb{Q}(\alpha) \subseteq \mathbb{Q}(\sqrt[3]{2})$. Now

$\sqrt[3]{2} + \sqrt[3]{4} \in \mathbb{Q}(\alpha)$

$(\sqrt[3]{4} + \sqrt[3]{2})^2 = 2 \sqrt[3]{2} + \sqrt[3]{4} + 4 \in \mathbb{Q}(\alpha)$

Subtracting the first element from the second implies $\sqrt[3]{2} \in \mathbb{Q}(\alpha)$, and therefore $\mathbb{Q}(\alpha) = \mathbb{Q}(\sqrt[3]{2})$.

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If $\alpha$ is rational, then so is $(2^{1/3} + 4^{1/3})$, and so the cube of that expression is rational; expanding, we see that the cube of that expression looks like a rational number plus $3*(32)^{1/3} = 6*4^{1/3}$; standard arguments tell you that this is irrational.

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How did you get this ? my calculations gives Wolfram result with both $2^{1/3}$ and $4^{1/3}$ ?. link:wolframalpha.com/input/?i=%282^%281%2F3%29%2B4^%281%2F3%29%29^3 –  Belgi Sep 9 '12 at 17:52
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Actually, if $r = 2^{1/3}$, $(\alpha - 1)^3 = 6 + 6 r + 6 r^2 = 6 \alpha$, so $\alpha$ is a root of the polynomial $(z - 1)^3 - 6 z = z^3 - 3 z^2 - 3 z - 1$, which has no rational roots. –  Robert Israel Sep 9 '12 at 17:56
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